Question

From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O

How many grams of O2 (32g/mol) will be needed to burn 36.1 g of B2H6 (Molar mass = 27.67g/mol)? ______g


Include the correct number of significant figures in your final answer

Answer 1

Answer: 125 g

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} B_2H_6=(36.1g)/(17)=1.30moles`

The balanced reaction is:

`B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O`

According to stoichiometry :

1 mole of
`B_2H_6` require = 3 moles of
`O_2`

Thus 1.30 moles of
`B_2H_6` will require=
`(3)/(1)* 1.30=3.90moles` of
`O_2`

Mass of
`O_2=moles* {\text {Molar mass}}=3.90moles* 32g/mol=125g`

Thus 125 g of
`O_2` will be needed to burn 36.1 g of
`B_2H_6`